本文共 3210 字，大约阅读时间需要 10 分钟。
The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-digit room numbers on their offices.
— It is a matter of security to change such things every now and then, to keep the enemy in the dark. — But look, I have chosen my number 1033 for good reasons. I am the Prime minister, you know! — I know, so therefore your new number 8179 is also a prime. You will just have to paste four new digits over the four old ones on your office door. — No, it’s not that simple. Suppose that I change the first digit to an 8, then the number will read 8033 which is not a prime! — I see, being the prime minister you cannot stand having a non-prime number on your door even for a few seconds. — Correct! So I must invent a scheme for going from 1033 to 8179 by a path of prime numbers where only one digit is changed from one prime to the next prime. Now, the minister of finance, who had been eavesdropping, intervened. — No unnecessary expenditure, please! I happen to know that the price of a digit is one pound. — Hmm, in that case I need a computer program to minimize the cost. You don't know some very cheap software gurus, do you? — In fact, I do. You see, there is this programming contest going on... Help the prime minister to find the cheapest prime path between any two given four-digit primes! The first digit must be nonzero, of course. Here is a solution in the case above.1033 1733 3733 3739 3779 8779 8179The cost of this solution is 6 pounds. Note that the digit 1 which got pasted over in step 2 can not be reused in the last step – a new 1 must be purchased.
31033 81791373 80171033 1033Sample Output
670
sprintf(s,"%d",n);
参考网址：
sscanf(s,"%d",&n);
从字符串s中取一个整数值到变量n中。
过程：每一位数字都从1-9枚举一次，如果是素数，就再次枚
举，直到找到目标答案。
#include#include #include #include #include #include #include using namespace std;const int N=10000;int prime[N];int s;int e;int vis[N];int Prime(int x){ for(int i=2;i<=sqrt(x);i++) { if(x%i==0) return 0; } return 1;}int change(int n,int k) // 把n的第k位转化为0； { char s[10]={0}; sprintf(s,"%d",n); //把整数n 打印成一个字符串保存在s 中 //参考网址：http://blog.sina.com.cn/s/blog_980cf62a0100ya0z.html s[k]='0'; sscanf(s,"%d",&n); // 从字符串s中取一个整数值到变量n中。 return n;}int bfs(int s, int e){ int q,w,r; memset(vis,0,sizeof(vis)); queue Q; Q.push(s); vis[s]=1; while(Q.size()) { s=Q.front(); Q.pop(); if(s==e) { return vis[s]-1; } w=1000; for(int i=0;i<4;i++) { q=change(s,i); for(int j=0;j<10;j++) { r=q+j*w; if(prime[r]==1 && vis[r]==0 ) { Q.push(r); vis[r]=vis[s]+1; } } w=w/10; } } return -1;}int main(){ int T; for(int i=1000;i<10000;i++) prime[i]=Prime(i); scanf("%d",&T); while(T--) { scanf("%d%d",&s,&e); //不要忘记加 & 符号 细节决定成败 if(bfs(s,e)==-1) { printf("Impossible\n"); } else { printf("%d\n",bfs(s,e)); } } return 0;}
转载地址：http://zioez.baihongyu.com/